Laplace Inverse

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February 19th 2013, 08:30 AM
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Laplace Inverse

Laplace Inverse of 1 is 1/s or Deta(t) I am confused , can anybody give me confirm answer

Thanks a lot in advance
February 19th 2013, 12:48 PM
HallsofIvy

Re: Laplace Inverse

Have you tried taking the Laplace transform of those?
The Laplace transform of f(x) is, by definition, $\int_0^\infty e^{-sx}f(x)dx$
So what are $\int_0^\infty (1)e^{-sx} dx$ and $\int_0^\infty (\delta(x))e^{-sx}dx$ ?
February 21st 2013, 06:22 AM
ssss

Re: Laplace Inverse

I know that Laplace of 1 is 1/s but I am not getting what is Laplace Inverse of 1. Also I am not clear about answer given by you

Thanks for the reply
February 21st 2013, 07:06 AM
HallsofIvy

Re: Laplace Inverse

Surely, you understand the relation between a "Laplace transform" and its inverse? (Or, for that matter, any kind of operation and it inverse!)

Yes, $L(1)= \int_0^\infty e^{-st}(1)dt= \left[-\frac{1}{s}e^{-st}\right]_0^\infty= \frac{1}{s}$ so 1 is the "inverse Laplace transform" of $\frac{1}{s}$ , not 1.

The Laplace transform of [tex]\delta(x)[\tex] is $L(\delta(x))= \int_0^\infty e^{-st}\delta(t)dt= e^0= 1$ .

Okat, NOW what is the inverse Laplace transform of "1"?
February 21st 2013, 07:15 AM
ssss

Re: Laplace Inverse

According to me its Delta function

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